SOLVED EXAMPLES
1. Distribute three particles in two different states according to (i) MB (ii) BE statistics.
Solution:
(i)
As per MB statistics particles are distinguishable, so total microstates
Ngini
23
as ni=3, N=3, gi=2
W
 3
ni
3
 (3  2 1) 
2 2 2
 8 microstates
3  2 1
Let P,Q,R are three particles then energy states 1,2 will have as below
Energy state 1
Energy state 2
PQR
0
PQ
R
P
Q
QR
PR
0
PQR
R
PQ
QR
PR
P
Q
So total microstates =8
(ii)
BE statistics
Here particles are indistinguishable, so
n
N i  gi  1
i 1
N i gi  1
W 

3  2 1
4

 4 microstates
3 2 1
31
As three particles are indistinguishable. Let us consider them as P. so
Energy state 1
Energy state 2
PPP
0
PP
P
0
PPP
P
PP
So total microstates = 4
(iii)
FD statistics
Here particles are indistinguishable and obey Pauli’s exclusion principle
W
gi
ni gi  ni

2
 ve which is not possible. So FD statistics can not be
3 23
applied on these particles.
2. Three distinguishable particles each of which can be accommodated in energy states E, 2E, 3E.
4E with total energy 6E. Find all the possible number of distributions. Also find total microstates
in each case.
(WBUT-2007).
Solution:
Suppose three particles are P,Q,R and distinguishable.
Possible
Arrangements
2,0,01
0,3,0,0
1,1,1,0
E
2E
3E
4E
PQ
PR
QR
0
P
Q
R
Q
R
P
0
0
0
PQR
Q
P
P
R
Q
R
0
0
0
0
R
R
Q
P
P
Q
R
Q
P
0
0
0
0
0
0
0
Total
Energy
6E
6E
6E
6E
6E
6E
6E
6E
6E
6E
Total
Microstates
3
1
6
Hence total arrangements = Macrostates = 3
And total microstates = 10
3. A system with non-degenerate single particle state with 0,1,2,3 energy units. Three particles are
to be distributed in three states so that the total energy of the system is 3 units. Find the number of
microstates if particles obey (i) MB statistics (ii) FD statistics.
(WBUT 2008).
Solution:
Let the particles be P,Q,R which are distinguishable. So for (i) MB statistics
Macrostates
0E
1E
2E
3E
0,3,0,0
2,0,0,1
0
PQ
PR
QR
P
Q
R
PQR
0
0
0
Q
P
P
0
0
0
0
R
R
Q
0
R
Q
P
0
0
0
1,1,1,0
Total
Energy
3E
3E
3E
3E
3E
3E
3E
Total
Microstates
1
3
6
Q
R
P
R
Q
R
P
P
Q
0
0
0
3E
3E
3E
So total microstates = 10
And for (ii) BE statistics- here particles are indistinguishable. suppose there are denoted by P
Macrostates
0E
1E
2E
3E
0,3,0,0
2,0,0,1
1,1,1,0
0
PP
P
PPP
0
P
0
0
P
0
P
0
Total
Energy
3E
3E
3E
Total
Microstates
1
1
1
So total microstates = 3
(iv)
FD statistics – here particles obey Pauli’s exclusion principle and they are
indistinguishable. So
Macrostates
0E
1E
2E
3E
Total
Total
Energy Microstates
1,1,1,0
P
P
P
0
3E
1
So total microstates = 1
4. Show that at T=0, the average energy E of an electron in a metal is given by E 
3
E f (0)
5
where Ef(0) is Fermi energy at absolute zero.
Solution:
As electron is a fermion. Total energy at absolute zero is given by
U (0)  
E f (0)
0
1
2

EN ( E )dE   f ( E ) ECE dE where C is the constant depending over other
0
properties.
And at absolute zero f(E) = 1 so
U (0)  C 
E f (0)
0

So
3
2 5
E 2 dE  C E f2 (0)
5
3
NE f (0)
5
3
E  E f (0)
5
hence proved
5. Calculate the Fermi energy in copper. Consider density of copper as 8.94 x 103 kg/m3 with atomic
mass 63.5 amu.
Solution:
Here electron density = N/V atom/m3
N N A  6.02 1026  8.94 103


V
W
63.5
=8.48 x 1028 atoms/m3
Fermi energy E f 
h  3 N 


2m  V 
2
2
1.05 10 

34 2
2  9.11031
 0.06 10
37
2
3
2
  3  9.86  8.48 1028  3
2
27 3
 2508.38 10 
 6.91 eV
6. Table shows the results for Fermi energies of some monovalent elements
Metal Cu
Li
Rb
Cs
Ag
K
Ef
7.04 4.72 1.82 1.53 5.51
2.12
If the Fermi velocity of electron in one of the metals is 0.73 x 106 m/s. Identify the metal. Also
calculate the Fermi temperature.
Solution:
At Fermi energy,
Kinetic Energy = E f 
1 2
mv f =0.5 x (9.1x10-31)(0.73x106)2
2
=2.42 x 10-19 joules
=1.51 eV
hence the metal is Cesium.
And Fermi energy at Fermi temperature, E f  kT f
Or
2.42 1019
Tf 

k
1.38 1023
Ef
= 1.75 x 104 K.
7. If the Fermi energy of any metal is 10 eV. What is the corresponding classical temperature?
Solution:
We have
E
3
3
kT  E f
2
5
So
T
2E f

5k
2 10 1.6 1019
5 1.38 1023
=4.64 x 104 K.
8. There are about 25 x 1028 free electrons/m3 in sodium. Calculate its Fermi energy, Fermi velocity
and Fermi temperature.
Solution:
Here
N
=electron density = 2.5 x 1028 /m3
V
34
h2  3N  3  6.62 10   3
3
Now Fermi energy E f 
 2.5 1028 

 
31 
2m  8V 
2  9.110  8

2
2
2
=5 x 10-19 Joules
=3.1 eV
This is the maximum kinetic energy of free electron at 0 K.
1
Then
1
E f  mv 2
2
And
E f  kT f
or
1
 2 E  2  2  5 1019  2
vf   f   
31 
 m   9.110 
=1.047 m/s
So
Tf 
5 1019
1.38 1023
= 3.623 x 104 K.
9. Calculate the extent of the energy range between f(E) =0.9 and f(E) = 0.1 at 200 K and express
the same as a function of Ef which is 3 eV.
Solution:
We know that
f E 
Ni
1

Ei  E f  / kT

gi 1  e

1
1 e

Where x 
(i)

 E f  E / kT

1
1  e x
Ef  E
kT
When f(E) = 0.9
so 0.9 
1
 0.9e  x  1  0.9  0.1
1  e x
1
or e x  9
9
So x  log e 9 =2.198
e x 
E f  E  2.198  kT
1.38 1023  200
Now kT 
eV
1.6 1019
=0.017 eV
When f(E) = 0.1
(ii)
1
on solving x  2.3026log10 9
1  ex
E1  E f
 2.3026  0.954
0.017
0.1 
E1 = Ef + 0.0.37 = 3 + 0.037 = 3.037 eV
So ΔE=E1-E = 3.037 – 2.963 = 0.074 eV
Or
E 0.074
= 0.025 = 2.5%

Ef
3
10. Calculate the temperature at which there is only one percent probability that a state, with energy
0.5 eV above Fermi energy will be occupied by an electron.
Solution:
We know that f  E  
1
1 e
 Ei  E f  / kT
Here E – Ef = 0.5 eV and f(E) = 1% = 1/100
So
1
1

100 1  e x
So
ex 
0.99
 99
0.01
Or
x  2.3026log10 99
Or
T
or
kT 
1.109 1.6 1019
= 1264 K
1.38 1023
0.5
 1.109 eV
2.3026 log10 99